Subject: Chemistry — Chemical Thermodynamics
Question: One mole of an ideal monoatomic gas undergoes two
reversible processes (A → B and B → C) as shown in the given figure:
[IMAGE] A → B is an adiabatic process. If the total heat absorbed in the
entire process (A → B and B → C) is
$RT_{2}\ln 10$, the value of $2\log V_{3}$ is
\_\_\_\_\_\_\_\_\_\_\_\_.
\[Use, molar heat capacity of the gas at constant pressure,
$C_{p,m} = \frac{5}{2}R$\]
Answer: (7)
Solution: A -- B (Adiabatic)
Monoatomic gas $\gamma = 5/3$
$$T_{1}V_{1}^{\gamma - 1} = T_{2}V_{2}^{\gamma - 1}$$
$$600V_{1}^{5/3 - 1} = 60V_{2}^{5/3 - 1}$$
$$10.V_{1}^{2/3} = V_{2}^{2/3}$$
$$10(10)^{2/3} = V_{2}^{2/3}$$
$v_{2} = 10^{5/2}$ ... (i)
$$q_{Total} = q_{A - B} + q_{B - C}$$
$$RT_{2}\ln 10 = 0 + RT_{2}\ln\frac{V_{3}}{V_{2}}$$
$$\frac{V_{3}}{V_{2}} = 10$$
$V_{3} = 10,V_{2} = 10^{7/2}$ from (i) put value of $V_{2}$
$$\log V_{3} = 7/2$$
$$2\log V_{3} = 2 \times \frac{7}{2} = 7$$